## Euler totient and Mobius function

### Euler totient and Mobius function

Hi,

Here is an identity to prove :

Is there a way to express mu(n) as function of phi(n)?

Thank you for any clue

Ps : I'm sorry I corrected the mistake.

Here is an identity to prove :

Is there a way to express mu(n) as function of phi(n)?

Thank you for any clue

Ps : I'm sorry I corrected the mistake.

Last edited by Alhazen on Fri Apr 06, 2012 8:35 pm, edited 1 time in total.

### Re: Euler totient and Mobius function

Is it a new result or is it yet known?

Thank you for any clue.

Thank you for any clue.

### Re: Euler totient and Mobius function

1: 1 0

2: 3 1

3: 6 1

I don't see them ever being equal. Why even sum the right side to n? floor(n/(n+1)) = 0 for all positive n, so any expression multiplied by that will be 0.

2: 3 1

3: 6 1

I don't see them ever being equal. Why even sum the right side to n? floor(n/(n+1)) = 0 for all positive n, so any expression multiplied by that will be 0.

### Re: Euler totient and Mobius function

You are right I did a big mistake.thundre wrote:1: 1 0

2: 3 1

3: 6 1

I don't see them ever being equal. Why even sum the right side to n? floor(n/(n+1)) = 0 for all positive n, so any expression multiplied by that will be 0.

It is k not k+1

Sorry sorry sorry.

### Re: Euler totient and Mobius function

The updated formula checks out.Alhazen wrote:It is k not k+1

a(n) seems to approach n

^{2}* .60792 for large n.

I have no idea how to prove the right and left sides are always equal.

### Re: Euler totient and Mobius function

Thank you very much and once again sorry my mistake.

### Re: Euler totient and Mobius function

ceil(n/k) is just floor(n/k) + 1 unless k divides n, so the right hand side is equal to:

$$\sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor \left(\left\lfloor \frac{n}{k} \right\rfloor + 1\right) - \sum_{k|n} \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor$$

$$= \sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor^2 + \sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor - \sum_{k|n} \mu\left(k\right) \left(\frac{n}{k}\right)$$

$$= S_1 + S_2 - S_3 $$

Now, we can prove the following:

S

S

S

Thus it is equal to your left hand side.

The proofs of these are relatively simple so you may try proving them yourself.

Hints:

The proof for S

The proof for S

$$\sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor \left(\left\lfloor \frac{n}{k} \right\rfloor + 1\right) - \sum_{k|n} \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor$$

$$= \sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor^2 + \sum_{k=1}^n \mu\left(k\right) \left\lfloor \frac{n}{k} \right\rfloor - \sum_{k|n} \mu\left(k\right) \left(\frac{n}{k}\right)$$

$$= S_1 + S_2 - S_3 $$

Now, we can prove the following:

S

_{1}= 2 [sum]_{1[le]k[le]n}φ(k) - 1S

_{2}= 1S

_{3}= φ(n)Thus it is equal to your left hand side.

The proofs of these are relatively simple so you may try proving them yourself.

Hints:

The proof for S

_{1}and S_{2}uses concepts from**Problem 351**(View Problem).The proof for S

_{3}uses the substitution l = n/k and Möbius inversion.
Last edited by kevinsogo on Thu Apr 12, 2012 1:09 am, edited 1 time in total.

### Re: Euler totient and Mobius function

Thank you for the proof.

No one found it in another forum.

Someone had found part of it.

No one found it in another forum.

Someone had found part of it.

### Re: Euler totient and Mobius function

And 0.60792 = 6/(π^2)thundre wrote:The updated formula checks out.Alhazen wrote:It is k not k+1

a(n) seems to approach n^{2}* .60792 for large n.

I have no idea how to prove the right and left sides are always equal.

### Re: Euler totient and Mobius function

6/(π^2) is the density of squarefree numbers.

What can we do with that?

What can we do with that?

- Lord_Farin
**Posts:**239**Joined:**Wed Jul 01, 2009 10:43 am**Location:**Netherlands

### Re: Euler totient and Mobius function

The squarefree numbers are the ones for which the Möbius function is nonzero.Alhazen wrote:6/(π^2) is the density of squarefree numbers.

What can we do with that?

### Re: Euler totient and Mobius function

Lord_Farin wrote:The squarefree numbers are the ones for which the Möbius function is nonzero.Alhazen wrote:6/(π^2) is the density of squarefree numbers.

What can we do with that?

Yes! You are right.

Did someone said otherwise?

There are some links to exploit.

Did you take a look to my new formulation of the Mertens function ?

### Re: Euler totient and Mobius function

From the Dirichlet convolution representation of the totient function

and because it requires knowing all the divisors of n.

Though it still sort of feels like I am cheating since I am still using lower order values of the mobius function,and because it requires knowing all the divisors of n.